If 2+i√3 is a root of the eqn x²+px+q=0, where p,q belongs to R, then find the ordered pair (p,q)
For any quadratic, complex roots or irrational roots always come as conjugate pairs
so if one root is 2 + i√3, the other must be 2 - i√3
sum of those roots = 4
product of those roots = 4 - 3i^2 = 7
In any quadratic of the form
x^2 + px + q = 0
the sum of roots = -p
the product of roots = q
so p = -4 and q = 7
(p,q) = (-4,7)